Pls kindly help to calculate the infinite sum of the below Sequence: (a*b^n)/n, a,b is constant, two n are natural number such as =1,2,3,4,5,6......., and 0
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Pls kindly help to calculate the infinite sum of the below Sequence:
(a*b^n)/n, a,b is constant, two n are natural number such as =1,2,3,4,5,6......., and 0<b<1
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- c++ Given two sequences with n integers between 0 and 9, interpreted as two n-digit integers, calculate a sequence of numbers that represents the sum of the two integers. Example: n = 8, 1st sequence 8 2 4 3 4 2 5 1 2nd sequence + 3 3 7 5 2 3 3 7 1 1 6 1 8 6 5 8 8Using foldl, define a function num2int :: [Int] -> Int -> Int that converts a list of numbers and the number n into an integer of base n. >num2int [3,4,5,6] 10 3456 >num2int [1,1,0,1,0,0,0,1,0,1,0,1,1,0] 2 3456 num2int :: [Int] -> Int -> Int num2int x y = foldl (\acc (a,b) -> ___a___ + b*y^a) 0 (___b___ (\x y ->(x,y)) [0..] (___c___ x)) fill in blank spaces of a , b , and cWrite a Python3 and C++ program for the given problem: Given two arrays X and Y of positive integers, find the number of pairs such that xy > yx (raised to power of) where x is an element from X and y is an element from Y. Example 1: Input: M = 3, X[] = [2 1 6] N = 2, Y[] = [1 5]Output: 3Explanation: The pairs which follow xy > yx are as such: 21 > 12, 25 > 52 and 61 > 16 .Example 2: Input: M = 4, X[] = [2 3 4 5]N = 3, Y[] = [1 2 3]Output: 5Explanation: The pairs for the given input are 21 > 12 , 31 > 13 , 32 > 23 , 41 > 14 , 51 > 15 . Your Task:This is a function problem. You only need to complete the function countPairs() that takes X, Y, M, N as parameters and returns the total number of pairs. Expected Time Complexity: O((N + M)log(N)).Expected Auxiliary Space: O(1). Constraints:1 ≤ M, N ≤ 1051 ≤ X[i], Y[i] ≤ 103
- Write a function that takes an unsigned integer andreturns the number of '1' bits it has(also known as the Hamming weight).For example, the 32-bit integer '11' has binaryrepresentation 00000000000000000000000000001011,so the function should return 3.T(n)- O(k) : k is the number of 1s present in binary representation.NOTE: this complexity is better than O(log n).e.g. for n = 00010100000000000000000000000000only 2 iterations are required.Number of loops isequal to the number of 1s in the binary representation."""def count_ones_recur(n):.Write a function that takes an unsigned integer andreturns the number of '1' bits it has(also known as the Hamming weight).For example, the 32-bit integer '11' has binaryrepresentation 00000000000000000000000000001011,so the function should return 3.T(n)- O(k) : k is the number of 1s present in binary representation.NOTE: this complexity is better than O(log n).e.g. for n = 00010100000000000000000000000000only 2 iterations are required..using c++ Given a positive integer, N, the ’3N+1’ sequence starting from N is defined as follows:If N is an even number, then divide N by two to get a new value for NIf N is an odd number, then multiply N by 3 and add 1 to get a new value for N.Continue to generate numbers in this way until N becomes equal to 1For example, starting from N = 3 the complete ’3N+1’ sequence would be:3, 10, 5, 16, 8, 4, 2, 1Write code to ask the user to enter a positive integer (N) in the main() function. Write a function sequence()that receives the integer value N and display the ‘3N+1’ sequence starting from the integer value that wasreceived (entered by the user). The function must also count and return the numbers that the sequenceconsists of. The returned value must be displayed from the main() function.
- Write a C++ program to perform addition of two hexadecimal numeralseach with up to 4 hex digits. If the result of the addition is more than 4 hex digitslong, then simply give the output message “Addition overflow” and stop. Use dynamicarrays to store hexadecimal numbers as arrays of characters. Assume the numbers canbe of different length (i.e. if adding hex numbers 12ab + 90 (first number is stored as 4characters in size, second number is stored as 2 characters in size), the second numberwill then be interpreted as 0090 but the leftmost zeros will not be included in the array(the size of the array variable will be useful here). To perform hex addition, you canconvert each individual hex digit to decimal, perform the addition, and then convert tohex, including adding a carry as necessary.To represent an integer value of n in decimal, we need to use ⌈log10(n + 1)⌉ digits . Verify this using n = 35, 1290. Note that for any real number x, ⌈x⌉ is the smallest integer that is greater than x, this means we always round up. For example, ⌈10.3⌉ = 11, ⌈−5.2⌉ = −5. To represent an intege value of n in binary, we need to use ⌈log2(n + 1)⌉ binary digits . Verify this using n = 35, 1290. Use the definition of logarithmic function to show that log2n / log10n = log210Write Algorithm to Conversion from an Arabic number to a modern Roman number.in: decimal number n (0 ≤ n)out: sequence R = s0, s1,...,s12 representing the structure of the Roman number (Ri = number of primitives Vi in n for i ∈ [0, 12])constant: sequence P = 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 of primitive Roman numberslocal: remainder x to be converted (0 ≤ x ≤ n); coefficient c for a primitive Romannumbers (for other than P0, 0 ≤ c ≤ 3)
- 1) For IEEE 754 single precision floating point numbers, if X is a negative denormalized floating point number with mantisa 010 0000 0000 0000 0000 0000, what is the decimal value of X? Question options: a. –1.25 × 2^(–126 ) b. –1.25 × 2^(–127 ) c. –0.25 × 2^(–127) d. –0.25 × 2^(–126 ) e. None of the above 2) Which of following recursive C++ functions correctly computes a[0]+a[1]+...+a[n-1] (where n is the size of array a) ? Question options: int sum(int n, int a[] ){ return n>0? sum(n-1) + a[n] : a[0]; } int sum(int n, int a[] ){ return n>0? sum(n-1, a) + a[n-1] : a[0]; } int sum(int n, int a[] ){ return sum(n-1, a) + a[n-1] ; } int sum(int n, int a[] ){ return n>0? sum(n-1, a) + a[n] : a[0]; } int sum(int n, int a[] ){ return n>0? sum(n-1) + a[n-1] : a[0]; }Consider the following code which includes a parity check for the third digit: C=(000,011,101,110). You get the third digit by adding the first two and then using the remainder on division by 2. Can this code detect and count single errors? how can this be implemented?A number of the form a + ib, in which i2 = -1 and a and b are real numbers, is called a complex number. We call the real part and b the imaginary part of a + ib. Complex numbers can also be represented as ordered pairs (a, b). The addition and multiplication of complex numbers are defined by the following rules: (a + ib) + (c + id) = (a + c) + i(b + d ) (a + ib) * (c + id) = (ac - bd) + i(ad + bc) Using the ordered pair notation, these rules are written as: (a, b) + (c, d) = ((a + c), (b + d )) (a, b) * (c, d) = ((ac - bd ), (ad + bc)) C++ has no built-in data type that allows us to manipulate complex numbers. Construct a data type, complex Type, that can be used to process complex numbers. Overload the stream insertion and stream extraction operators for easy input and output. We will also overload the operators + and * to perform addition and multiplication of complex numbers. If x and y are complex numbers, we can evaluate expressions such as x + y and x * y.